IB Maths AA HL · Paper 1

Another way. Expand to $3x^2-3(2+q)x+6q$ and set $-\frac{b}{2a}=4$. It gives the same result.

1. Displacement is the integral of velocity, so start from your answer to (a).
2. Starting at $O$ means $s=0$ when $t=0$. Substitute to find $C$.

Another way. Integrate the velocity from $0$ to $t$ as a definite integral. The constant takes care of itself.

1. Two terms apart means multiplying by $r$ twice, so $u_8=u_6{r}^2$.
2. Divide to get $r^2$, take the square root, and use the fact that every term is positive to choose the sign.

1. Change the base from $25$ to $5$, using ${\log }_{5}25=2$.
2. Write $81x^2$ as $(9x{)}^2$, then use the power law and the product law.

Another way. Let ${\log }_{25}(81x^2)=y$, so $25^y=81x^2$. That is $(5^y{)}^2=(9x{)}^2$, so $5^y=9x$ and $y={\log }_5(9x)$.

1. Replace ${\log }_{25}(81x^2)$ using part (a).
2. Gather the log terms in $x$ on one side and the numbers on the other, then drop the logs.
3. Undo the power $\frac{2}{3}$ by raising both sides to $\frac{3}{2}$.

1. Find $P(B)$ the same way: $B$ splits into $A\cap B$ and $A^{\prime }\cap B$.
2. Independence means $P(A\cap B)=P(A)\times P(B)$. Set that up.
3. Clear the fractions, collect into a quadratic and factorise.

1. Removing a sector of angle $\frac{\pi }{3}$ leaves a sector of angle $2\pi -\frac{\pi }{3}=\frac{5\pi }{3}$ in both circles.
2. Shaded area: the small sector of radius $6$. Unshaded area: the big sector minus the small one.
3. Set up the ratio of $\frac{2}{3}$. The common factor $\frac{1}{2}\cdot \frac{5\pi }{3}$ cancels, leaving a simple equation in $r^2$.

Another way. The shaded part is $\frac{3}{5}$ of what remains, so $36=\frac{3}{5}{r}^2$ straight away, giving $r^2=60$.

1. A positive result can happen two ways: pregnant and positive, or not pregnant and positive. A tree diagram lays this out.
2. Multiply along each branch and add the two results.

1. Use $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ with $A$ = pregnant and $B$ = positive.
2. The top is the pregnant-and-positive branch from (a); the bottom is your answer to (a).

1. The coefficients are real, so complex roots come in conjugate pairs. The five roots are $\frac{1}{3}$, $a\pm \mathrm{i}$ and $2\pm b\mathrm{i}$.
2. (i) Sum of roots $=-\frac{-7}{3}=\frac{7}{3}$. The imaginary parts cancel.
3. (ii) Product of roots for degree $5$ is $(-1{)}^5\times \frac{-26}{3}=\frac{26}{3}$. Each conjugate pair multiplies to a real number.
4. Use $a=-1$ and solve for $b$, keeping the value with $b>1$.

1. The real root $\frac{1}{3}$ gives a linear factor. Scale it to $3z-1$ so the leading coefficient $3$ is accounted for.
2. Each conjugate pair $p\pm q\mathrm{i}$ gives the quadratic $z^2-2pz+(p^2+q^2)$.

1. Dot $\mathbf{v}$ with $\mathbf{a}$. The brackets $(\mathbf{a}\cdot \mathbf{c})$ and $(\mathbf{a}\cdot \mathbf{b})$ are just numbers, so they come outside.
2. The scalar product is commutative, so the two terms are equal and cancel.

Another way. $\mathbf{v}$ is a combination of $\mathbf{b}$ and $\mathbf{c}$, so it lies in their plane, and $\mathbf{b}\times \mathbf{c}$ is normal to that plane.

1. From (a) and (b), $\mathbf{v}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}\times \mathbf{c}$.
2. A cross product of two vectors is perpendicular to both of them, so $\mathbf{a}\times (\mathbf{b}\times \mathbf{c})$ is perpendicular to the same pair. In 3D, two vectors perpendicular to the same two non-parallel vectors point along the same line.

1. Differentiate twice.
2. At a point of inflexion the second derivative is zero: $f^{\prime \prime }(-2)=0$ gives $b=6a$.
3. The point lies on the curve: substituting $x=-2$ into $f$ gives $-2a+b=6$.
4. Solve the two equations together.

1. $P$ is where the graph meets the $y$-axis, so $P=(0,d)$.
2. The tangent at $x=-3$ is horizontal, so it is the line $y=f(-3)$. Show that $f(-3)=d$.

1. Start from the right-hand side and write everything in $\tan \theta$, using $\frac{1}{\cot \theta }=\tan \theta$.
2. Combine over a common denominator and recognise the double angle formula for $\tan 2\theta$.

Another way. Write the right-hand side in $\sin$ and $\cos$: it becomes $\frac{{\cos }^2\theta -{\sin }^2\theta }{2\sin \theta \cos \theta }=\frac{\cos 2\theta }{\sin 2\theta }$.

1. Replace $\cot 2\theta$ using (a). Multiply through by $2\cot \theta$ (allowed, as $\cot \theta \ne 0$ when $\theta \ne \frac{\pi }{2}$).
2. This gives a quadratic in ${\cot }^2\theta$. Do not cancel ${\cot }^2\theta -1$, or you lose roots.
3. Factorise, find all values of $\cot \theta$, then list the angles in $(0,\pi )$.

1. Left side: the chain rule on $\cot 2\theta$ gives $-2{\mathrm{cosec}}^{2}2\theta$.
2. Right side: rewrite $\frac{1}{\cot \theta }$ as $\tan \theta$ first, then differentiate each term.
3. Set the two derivatives equal and multiply by $-2$.

1. The right angle is at $P$, so $QR$ is the hypotenuse. Find $PQ$ with Pythagoras and the identity from (c).
2. Area is half the product of the two shorter sides, $PR$ and $PQ$.
3. Use $\sin 2\theta =2\sin \theta \cos \theta$ to write the area in terms of $\mathrm{cosec}2\theta$, then compare with $L$.

1. Factorise: $-3+4x-x^2=-(x-1)(x-3)=(x-1)(3-x)$.
2. Write $1=A(3-x)+B(x-1)$ and substitute $x=1$ and $x=3$.

1. Separate the variables and use (a) on the left.
2. Integrate. Careful with $\frac{1}{3-N}$: the minus sign inside gives $-\ln (3-N)$.
3. Combine the logs and double everything, then add one constant.
4. Initially $1500$ rabbits means $N=1.5$ (thousands) at $t=0$. Substitute to find the constant.

1. Exponentiate both sides.
2. Cross-multiply, collect the $N$ terms on one side and factorise.
3. Divide top and bottom by ${\mathrm{e}}^{2kt}$ to reach the required form.

1. As $t\to \infty$, ${\mathrm{e}}^{-2kt}\to 0$ because $k>0$.
2. $N$ is in thousands, so convert back to a number of rabbits.